How do you integrate #int (1-x^2)/((x-9)(x+6)(x-4)) # using partial fractions?

1 Answer
Sep 10, 2016

#int (1-x^2)/((x-9)(x+6)(x-4)) dx =#

#=-16/15 ln abs(x-9) - 7/30 ln abs(x+6) + 3/10 ln abs(x-4) + C#

Explanation:

#(1-x^2)/((x-9)(x+6)(x-4)) = A/(x-9)+B/(x+6)+C/(x-4)#

We can determine #A#, #B# and #C# using Heaviside's cover up method:

#A = (1-color(blue)(9)^2)/((color(blue)(9)+6)(color(blue)(9)-4)) = (-80)/((15)(5)) = -16/15#

#B = (1-color(blue)((-6))^2)/((color(blue)((-6))-9)(color(blue)((-6))-4)) = (-35)/((-15)(-10)) = -7/30#

#C = (1-color(blue)(4)^2)/((color(blue)(4)-9)(color(blue)(4)+6)) = (-15)/((-5)(10)) = 3/10#

So:

#int (1-x^2)/((x-9)(x+6)(x-4)) dx =#

#int -16/15(1/(x-9)) - 7/30(1/(x+6))+3/10(1/(x-4)) dx#

#=-16/15 ln abs(x-9) - 7/30 ln abs(x+6) + 3/10 ln abs(x-4) + C#

where #C# is the constant of integeration (not to be confused with the coefficient #C# we calculated earlier).